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User blog:KthulhuHimself/Ordinal Arithmetic Notation (OAN)
Been brewing in my head for some time, so let's get it on paper: Introduction to Ordinal Arithmetic There's a transfinite ordinal I like named ε₀ (epsilon zero). It's defined as the limit to the sequence (ω, ω^ω, ω^(ω^ω),...), and can be thought of as ω^^ω. Now, when I saw the ordinal ω^^ω, I thought "why not go further?", and proudly wrote ω^^(ω+1) on a piece of paper. However, once I tried defining it, to my dismay I realised that under the conventional definition of tetration, it would be equal to ω^(ω^^ω). epsilon zero, being the fixed point for ω^a, is also equal to ω^(ω^^ω)! That's no good. So I quickly searched for an alternative definition. Although unintuitive at first, defining ω^^(ω+1) = (ω^^ω)^^ω served my purpose well... Of generalising arrow notation to transfinite ordinals. Straightforward definition Let a be some ordinal, b be some transfinite limit ordinal (other than omega, where b is the limit of sequence s), b+n some transfinite successor ordinal, n be an integer and ^k be our arrow notation: a^k+1n = a^k(a^k...) (n deep) a^k+1ω = limit of a^k(a^k...) a^kb = limit of a^ks a^k(b+(n+1)) = (a^kb+n)^kb (for k = 0,1,2; we treat a^kb as regular ordinal addition, multiplication and exponentiation respectively) So there you have it. Things like ω^^^^^(ω^^^4) are now not only well-defined, but useful for googology. Let's briefly expand this for transfinite numbers of arrows, where k is transfinite, limit of sequence s: a^kb = limit of a^sb And we're done. We've got ordinals like w^ww on our table now. Hyperoperators of transfinite order involving finite numbers as arguments Let's ask ourselves, what happens when we bring some integer n to integer m, by a hyperoperator of transfinite order? For example, what is 3^ω+164 equal to? Do I really need to explain it? I mean, isn't it already obvious? Let a be limit of a sequence, each member expressed as s(k) (kth member of the sequence), and a+n some successor ordinal: n^0m = n+m n^am = n^s(m)n n^a+(n+1)m = n^a+n(n^a+n...) (nested m times) And that's it. We've defined a straightforward notation, and a fairly powerful one at that. Analysing the notation so far The best way to explain this notation is to demonstrate some examples of it. 3^33 = 3^2(3^23) = 3^227 = 7,625,597,584,987 5^43 = 5^3(5^35) = 5^3(5^2(5^2(5^2(5^25)))) = 5^3(5^2(5^2(5^2(5^1(5^1(5^1(5^15))))))) = 5^3(5^2(5^2(5^23125))) > 5^3(5^2(5^2(10^2184))) > 5^3(10^(10^(10^2184))) = 5^2(5^2(5^2...)) nested 10^(10^2184) times = 5^^(10^(10^2184)) But clearly these are small fries. Remember 3^ω+164? 3^ω+165 is already larger than Graham's number. STILL small fries. How about something crazy, like 3^[ω^ωω]3? It's apparent that in general, n^an is more or less equivalent to fa(n) in FGH, so to analyse larger numbers in this notation, we only need to analyse the ordinal in the hyperoperator. So what is ω^ωω? by the definition above, it's the limit of the sequence ω^0ω, ω^1ω, ω^2ω, ω^3ω,... etc. So let's analyse each of those. ω^0ω = ω*2 ω^1ω = ω2 ω^2ω = ωω ω^3ω = limit of ωωωω..., or in other words, ε₀ ω^4ω = here we find our first really quite interesting ordinal. Clearly it's the limit of ω^3(ω^3...), but what is ω^3(ω^3ω) or in other words, ω^^ε₀? It's the limit of ω^3(ω^31), ω^3(ω^32), etc. Let's briefly analyse this sequence. ω^3(ω^31) = ω^3ω = ε₀ ω^3(ω^32) = ω^3(ω^2ω) = ω^3(ω2) = the limit of another fundamental sequence (ahead of time I'm guessing it's εω2). To analyse ω^^(ω^2), let's first talk about ω^^(ω+1). According to our definition, it's the same as (ω^^ω)^^ω, which is the same as ε₀^^ω. This is obviously the first fixed point of a -> ε₀^a, which is the definition of ε₁! It isn't hard to see that ω^^(ω+2) is the next member of the epsilon series, ω^^(ω*2) is the ωth member, ω^^(ω*(n+1)) is the (ω*n)th member and ω^^(ω^2) is the (ω^2)th member, or εω2, as I predicted. Now, given this definition, it's clear that ω^4ω is the first fixed point of εa, meaning that ω^4ω is equivalent to zetta_0. I think it's pretty clear that the way ω^kω is defined, ordinals for specific k are equal to φ(0,k), which extends to transfinite ordinals. In other words, the current system is limited by the Feferman-Schütte Ordinal (Γ0), with the fixed point of ω^[ω^[ω^...ω]ω]ω. So it's time for our first extension. Transfinite Hyperoperator Array Notation (THAN) In order to create this notation, we'll first need to define a few new ideas regarding these ordinal hyperoperators, the first of which being a "stack" of operators. We'll notate ω{n} to describe ω^[ω^[ω^...ω...ω]ω]ω nested n times. We'll call this a "stack of omega with operators n pieces high". Defining ω{ω} is intuitive, as the limit of ω{n} for natural numbers (coincidentally, ω{ω} is equal to Gamma_Zero), but what does ω{ω+1} equal to? As with ω^^(ω+1), the same problem is present and the definition which solves it is fairly straightforward, and in fact very similar. We'll have ω{ω+1} = (ω{ω}){ω}. That is, it is a "stack of ω{ω} with operators omega pieces high", that is, the limit of all finite stacks of ω{ω}. Given this definition, we can expand our system to all countable ordinals with ease. Let a be an ordinal, n an integer, b a limit ordinal of sequence s and b+n some successor ordinal: a{n} = stack of the ordinal with operators n pieces high. a{b} = limit of a{s} a{b+(n+1)} = (a{b+n}){b} And that's it! Now that we have a good definition for ω{a}, with any ordinal a given, the next step is intuitive, since we can nest ordinals of the ω{a} sort into a. In fact, here comes our next entry into the array I've been promising. But first, we should define a little concept named "recursion pattern". In fact we've been using it throughout this entire blog post without naming it explicitly. When we say that function f(k) is nested in k by transfinite recursion pattern a, we really mean that by using the fast growing hierarchy, if we define F0(k) as f(k), then nesting it "by a" means we're talking about Fa(k). With that in mind, let's expand our notation into an array notation. (remember that a is the transfinite limit of sequence s) ω{0,1} = ω{a} nested by a recursion pattern of ω ω{b+(n+1),1} = ω{a} nested in a by a recursion pattern of ω{(b+n),1} ω{a,1} = limit of ω{s,1} ω{0,2} = ω{a,1} nested in a by a recursion pattern of ω In general: ω{0,a} = limit of ω{0,s} ω{0,b+(n+1)} = ω{a,b+n} nested in a by a recursion pattern of ω ω{c+(n+1),b+(n+1)} = ω{a,b+n} nested in a by a recursion pattern of ω{c+n,b+(n+1)}. Now, things get even more insane if we extend this notation to longer arrays by treating the two entries in the above definition as the last two entries in a longer array. It's easy to demonstrate that this extended definition works, by observing that whenver the last entry is one, the ordinal is defined according to an array one entry shorter, ensuring that every ordinal defined by a long array collapses down to shorter and shoter arrays until its definition includes something more standard. So yeah, these are big ordinals. It'll take me some time to figure out just how huge they are, but I'll post an analysis soon enough. There's one last extention I'd like to add to it, involving arrays of transfinite length, but tell me you; don't you think the numbers this notation could provide would be massive? Just try to work out in your head what's the size of 10^ω{100}100 (the thangol), 5^ω{0,0,2}5 (the megapentacle), or 3^ω{0,0,0,0,0,1}3 (the amazing trinormous-six)! Transfinite Number of entries in THAN Now, for transfinite ordinals, we'll denote places on the array using the following notation: ω{...x(a)y...} = ω{...x,0,0,0,...a entries...,y...} So what is ω{0(ω)1} equal to? It should be quite obvious, that it is the limit of the sequence ω{0,1}, ω{0,0,1}, ω{0,0,0,1}..., and being a limit ordinal for a fundamental squence, we can use it in our original definition of THAN. Following this logic, let's analyse some of the next ordinals this new extension gives us: ω{1(ω)1} = stack of the ordinal ω{0(ω)1} with operators stacked 1 high. ω{n(ω)1} = stack of the ordinal ω{0(ω)1} with operators stacked n high. Small clarifying note When I write ω^^ω, I mean ω^3ω. for k arrows, the notation I use is ω^k+1ω (k smaller than ω). Category:Blog posts